Question

The equation $f(x)=7a^{2+x}-b$ has an x-intercept equivalent to


A) $x=\frac{logb-log7}{loga}-2$

B) $x=7a^{2}-b$

C) $x=\frac{y+b}{7a^{2}}$

D) $x=0$

I have the answer key so I know the answer. I just don't know how they got to the answer so please show your work.

Answer 1

Answer:

$\textsf{A)} \quad x=\dfrac{\log b - \log 7}{ \log a}-2$

Step-by-step explanation:

Given function:

$f(x)=7a^{2+x}-b$

The x-intercepts occur when the function equals zero.

$\implies 7a^{2+x}-b=0$

Add b to both sides:

$\implies 7a^{2+x}=b$

Divide both sides by 7:

$\implies a^{2+x}=\dfrac{b}{7}$

Take logs of both sides of the equation:

$\implies \log \left(a^{2+x}\right)= \log \left(\dfrac{b}{7}\right)$

$\textsf{Apply the Power log law}: \quad \log_ax^n=n\log_ax$

$\implies (2+x)\log a= \log \left(\dfrac{b}{7}\right)$

$\textsf{Apply the Quotient log law}: \quad \log_a\frac{x}{y}=\log_ax - \log_ay$

$\implies (2+x)\log a= \log b-\log 7$

Divide both sides by log a:

$\implies 2+x=\dfrac{\log b - \log 7}{ \log a}$

Subtract 2 from both sides:

$\implies x=\dfrac{\log b - \log 7}{ \log a}-2$